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# Profit & Loss Tricks & Tips

Profit and  Loss is yet another crucial topic of Arithmetic section of Quantitative Aptitude. You may also find the application of this topic in certain Data Interpretation questions as well. To help you all prepare for this topic better, we will discuss about this topic today. Before going to detail in this topic, we recommend you to go through the following topics -Percentage and Ratio

Profit and loss problems involve various terms like cost price, selling price, marked price etc. Basically, it is a difference between selling price and cost price.

Cost price is the price paid to purchase an article or a product or we can say it is a cost incurred in manufacturing an article.

Selling price is the price at which a product is sold.

### Profit and loss formulas used in profit and loss:

1) Generally, profit is calculated as:
Profit or gain = Selling price(S.P) – Cost price(C.P)

#### SP Formula

⇒ Selling Price = Cost Price + Profit

2)  Similarly,  Loss = Cost price – Selling price

3)  Gain percentage(%)

4)  Loss percentage(%)

5) There is a direct relationship between selling price and cost price:

For Example: If an article is sold at a gain of 27%, then by using the first formula, you can find that S.P. is 127% of C.P.
Similarly, If an article is sold at loss of 18%, then by using the second formula, you can find that S.P. is 82% of C.P.

6)  If a person sells two commodities at same prices. On one he gains x% and loses x% on another, then as a whole he will be in loss and the loss percentage will be equal to:

Note: Here is an example to find gain in case of dishonesty.
Problem 1: A dishonest dealer professes to sell his goods at cost price but he uses a weigh 960 grammes for 1 kg. How to calculate the gain percentage?
Solution:

Here I am providing a basic concept of Profit And Loss Problem Shortcut Tricks and Formulas Which will help you scoring good marks in the exam and also save your valuable time.

Marked price(M.P):The normal price of the thing without any discount.

Cost price(C.P):The price at which an article is purchased is called its cost price.

Selling price(S.P):The price at which an article is sold is called its selling price.

Profit or Gain:If S.P>C.P, The seller is said to have a Profit or Gain.

Loss:If S.P<C.P, The seller is said to have a Loss.

Concept:Discount% always calculated on marked price. and profit% and loss% always calculate on Cost Price(C.P)

Profit And Loss Problem Shortcut Tricks And Formulas

Profit or Gain=S.P-C.P

Loss=C.P-S.P

%Profit or Gain= (Profit or Gain×100)/C.P

%Loss=(Loss×100)/C.P

Concept:

Gain or Profit of x% can be represented as (100+x)/100

Loss of x% can be represented as (100-x)/100

If C.P and Profit% or Gain% given then

S.P=C.P×(100+Gain%)/100

If C.P and Loss’s% given then

S.P=C.P×(100-Loss%)/100

If S.P and Profit% or Gain% given then

C.P=(S.P×100)/(100+Gain%)

If S.P and Loss’s% given then

C.P=(S.P×100)/(100-Loss%)

Common Type Of Profit and Loss Problem Shortcut Tricks That Are Frequently Asked In Various Competitive Exam

For Example:

(Q1)A man buys a cycle for Rs. 1400 and sells it at a loss of 15%. What is the selling price of the cycle?

Solution:

=>1400×(100-15)/100=S.P

=>1400×(85/100)=S.P

S.P=Rs. 1190 Ans.

(Q2)A man buys a cycle for Rs. 1400 and sells it at a profit of 20%. What is the selling price of the cycle?

Solution:

=>1400×(100+20)/100=S.P

=>1400×(120/100)=S.P

S.P=Rs. 1680 Ans.

(Q3)There is 10% loss if an article is sold at Rs. 270, then the cost price of the article is.

Solution:

=>x×(100-10)/100=270

=>x×(90/100)=270

∴ x=(270×100)/90

=Rs. 300 Ans.

BY Formula:

C.P=(S.P×100)/(100-Loss%)

C.P=(270×100)/(100-10)

=(270×100)/90

=Rs. 300 Ans.

(Q4)An item when sold for Rs. 1690 earned 30% profit on the cost price. Then the cost price is.

Solution:

=>x×(100+30)/100=1690

=>x×(130/100)=1690

∴ x=(1690×100)/130

=Rs. 1300 Ans.

BY Formula:

C.P=(S.P×100)/(100+Profit%%)

C.P=(1690×100)/(100+30)

=(1690×100)/130

=Rs. 1300 Ans.

(Q5)A man buys an article for Rs. 27.50 and sells it for Rs. 28.60.Find his gain percent.

Solution:C.P=Rs. 27.50

S.P=Rs. 28.60

Gain=S.P-C.P

=Rs. 28.60-Rs. 27.50

=Rs. 1.10

∴ Gain%= (Profit or Gain×100)/C.P

= (1.10×100)/27.50

= 4% Ans.

(Q6)If a radio is purchased for Rs. 490 and sold for Rs. 465.50, find the loss percent.

Solution:C.P=Rs. 490

S.P=Rs. 465.50

Loss=C.P-S.P

=Rs. 490-Rs. 465.50

=Rs. 24.50

∴ Loss%=(Loss×100)/C.P

=(24.50×100)/490

=5% Ans.

(Q7)A fan is listed at Rs. 150 and a discount of 20% are given. Then the selling price is.

Solution:M.P=Rs. 150 Discount%=20%

Discount=150×(20/100)

=Rs. 30

S.P=M.P-Discount

=Rs. 150-Rs. 30

=Rs 120 Ans.

Another way:

=>150×(100-20)/100=S.P

=>150×(80/100)=S.P

S.P=Rs 120 Ans.

(Q8)If a Book costs Rs. 64 after 20% discount is allowed. What was its original price(i.e M.P) in Rs. ?

Solution:

=>x×(100-20)/100=64

=>x×(80/100)=64

∴ x=(64×100)/80

= Rs 80 Ans.

(Q9)A merchant purchases a table for Rs 450 and fixes its list price in such a way that after allowing a discount of 10%, he earns a profit of 20%. Then the list price of the table is.

Solution:C.P=Rs 450

Discount=10% and Profit=20%

M.P=?

=>450×(100+20)/100=S.P

=>450×(120/100)=S.P

S.P=Rs 540

=>x×(100-10)/100=540

=>x×(90/100)=540

∴ x=(540×100)/90

=Rs 600 Ans.

(Q10)Mohan purchased a machine for Rs 80,000 and spent Rs 5000 on repair and Rs 1000 on transport and sold it with 25% profit. At what price did he sell the machine?

Solution:Total cost price=80,000+5000+1000

=Rs 86,000

S.P=86000×(100+25)/100

=86000×(125)/100

=Rs 107500 Ans.

(Q11)An Egg seller buys 240 egg for Rs. 600.Some of these eggs are bad and are thrown away.He sells the remaining egg at Rs. 3.50 each and makes a profit of 198. The percent of an egg thrown away are.

Solution:C.P of an egg=600/240

=Rs. 2.5

profit=Rs 198

Total S.P=Profit+C.P     [where Profit or Gain= S.P-C.P ]

= 198+600=Rs 798

Number of egg sold=798/3.5

=228

Now,

Required percentage=(12×100)/240

= 5% Ans.

(Q12)A shopkeeper bought 30Kg of rice at the rate of Rs. 70 per kg and 20 Kg of rice at a rate of Rs. 70.75 per Kg. If he mixed the two brands of rice and sold the mixture at Rs. 80.50 per Kg, his gain is.

Solution:

Gain=S.P-C.P

= 4025-3515=Rs 510 Ans.

(Q13)A profit of 12% is made when a mobile phone is sold at Rs P and there is 4% loss when the phone is sold at Rs Q. Then Q:P is.

Solution:

=>x×(100+12)/100=P

=>x×(112/100)=P

112x=100P

P=112x/100

Now,

=>x×(100-4)/100=Q

=>x×(96/100)=Q

=>96x=100Q

Q=96x/100

Now,

Q/P=(96x/100)/(112x/100)

=6/7

Q:P=6:7 Ans.

(Q14)If the ratio of cost price to selling price is 10:11, Then the rate of a percent of profit is.

Solution:Let C.P=10x

S.P=11x

Profit=11x-10x=x

∴ % Profit=(x×100)/10x

=10% Ans.

(Q15)An article is sold at 5% profit. The ratio of selling price and cost price will be.

Solution:

=>x×(100+5)/100=S.P

S.P=105x/100

Now,

S.P/C.P=(105x/100)/x

=21/20

∴ S.P:C.P=21:20 Ans.

The common and frequently asked a question of Profit and Loss Problem Shortcut Tricks for Bank PO, SSC CGL, SSC CHSL, Railway Exam Type-2.

For Example:

(Q1)If a man were to sell his chair for Rs. 720, he would loss 25%.To gain 25% he would sell it for.

Solution:

Step-1

100x×(100-25)/100=720

75x=720

∴ x=720/75

C.P=100x=100×(720/75)=Rs. 960

Step-2

960×(100+25)/100=S.P

960×(125/100)=S.P

S.P=Rs. 1200 Ans.

BY TRICK:

Desire Gain(D Gain)=25%

Loss=25%

S.P=Rs(720×(100+D Gain))/(100-loss)

=Rs(720×125)/75

=Rs. 1200 Ans.

(Q2)By selling an article for Rs 600 a man gains 20%. At what price should he have sold it to gain 15%?

Solution:

Step-1

100x×(100+20)/100=Rs 600

120x=Rs 600

x=5

∴ C.P=100x=100×5

=Rs 500

Step-2

500×(100+15)/100=S.P

S.P=Rs 575 Ans.

BY TRICK:

Desire Gain(D Gain)=15%

Normal Gain=20%

S.P=Rs(600×(100+D Gain))/(100+Gain)

=Rs(600×115)/120

=Rs. 575 Ans.

(Q3)By selling an article for Rs 720 a man loss 20%. At what price should he have sold it to loss 15%?

Solution:

Step-1

100x×(100-20)/100=720

80x=720

∴ x=720/80=9

C.P=100x=100×9=Rs 900

Step-2

900×(100-15)/100=S.P

900×(85/100)=S.P

S.P=Rs 765 Ans.

BY TRICK:

Desire Loss(D Loss)=15%

Loss=20%

S.P=Rs(720×(100-D Loss))/(100-Loss)

=Rs(720×85)/80

=Rs 765 Ans.

(Q4)By selling 80 ball pens for Rs 140 a retailer loses 30%. How many ball pens should he sell for Rs 104 so as to make a profit of 30%?

Solution:

Step-1:

100x×(100-30)/100=140

70x=140

x=2

∴ C.P=100x=100×2=Rs 200

Step-2:

200×(100+30)/100=S.P

S.P=Rs 260

Now,

Rs 260=80 ball pen

∴ Rs 104=(80/260)×104

=32 ball pens Ans

BY TRICK:

Desire Profit(D-Profit)=30%

Loss=30%

S.P=Rs(140×(100+D Gain))/(100-loss)

=Rs(140×130)/70

=Rs. 260

Now,

Rs 260=80 ball pen

∴ Rs 104=(80/260)×104

=32 ball pens Ans

The common and frequently asked a question of Profit and Loss Problem Shortcut Tricks for Bank PO, SSC CGL, SSC CHSL, Railway Exam Type-2.1.

(Q1)A man sells a car to his friend at 10% loss. If the friend sells it for Rs 54,000 and gains 20% the original cost price of the car was.

Solution:

Let Original Cost Price or C.P of a Man=100x

100x×((100-10)/100)×(100+20)/100=54000

=>100x×(90/100)×120/100=54000

⇒108x=54000

∴ x=500

Original C.P=100x=100×500

=Rs 50000 Ans.

(Q2)A sold an article to B at the profit of 25%.B sold it to C loss of 10% and C sold it to D at the profit of 20%.If D paid Rs 27 then how much A paid to buy the article.

Solution:

Let C.P of A=100x

100x×(125/100)×(90/100)×(120/100)=Rs 27

∴ x=(27×100×100×100)/(100×125×90×120)

=0.2

C.P of A=100x=100×0.2

=Rs 20 Ans.

(Q3)A sold a horse to B for Rs 4800 by losing 20%.B sells it to C at a price which would have given A a profit of 15% B’s gain is.

Solution:

Let C.P of A=100x

100x×(80/100)=4800

80x=4800

∴ x=60

C.P of A=100x=100×60

=Rs 6000

Now,

6000×(115/100)=S.P of A

S.P of A(i.e C.P Of C)=6900

Gain of B=6900-4800

=Rs 2100 Ans.

(Q1)A vendor bought bananas at 6 for Rs 10 and sold them at 4 for Rs 6.Find his gain or loss percent.

Solution: By Concept And Shortcut Tricks

Step-2:

C.P=40 and S.P=36

Loss=C.P-S.P=40-36=Rs 4

% Loss=(4×100)/40

=10% Ans.

CONCEPT:[इस Type के Problem को solve करते Time Object को Only Equal बनाना है| एवं Object को equal बनाने के लिए जिस Number से Multiply किया जायेगा उसी Number से Rate वाले part को Multiply करना है]

Note: Above वाले method से आप लोग इस तरह के problem को 10 sec में solve कर सकते है और Formula याद रखने के कोई जरूरत नहीं है|

(Q2)A man buys lemons at the rate of 3 for Rs 5 and sells them at 2 for Rs 4 what is the profit percent?

Solution: By Concept And Shortcut Tricks

C.P=Rs 10 and S.P=Rs 12

Profit=S.P-C.P

=Rs(12-10)=Rs 2

% profit=(2×100)/10

=20% Ans.

(Q3)A man purchases some oranges at a rate of 3 for Rs 40 and same quantity at 5 for Rs 60. If he sells all the oranges at the rate of 3 for Rs 50, find his gain or loss percentage.

Solution: By Concept And Shortcut Tricks

Now,

C.P=Rs 380 And S.P=Rs 500

Profit=S.P-C.P

=Rs(500-380)

=Rs 120

% profit=(120×100)/380

=32% Ans.

The common and frequently asked a question of Profit and Loss Problem with solution and Shortcut Tricks for Bank PO, SSC CGL, SSC CHSL, Railway Exam Type-3.1

(Q1)The cost price of 15 articles is same as the selling price of 10 articles. The profit percent is.

Solution: By Concept And Shortcut Tricks

C.P of 15 articles = S.P of 10 articles

C.P/S.P=10/15

Profit=S.P-C.P

=15-10=5 articles

∴ Profit %=(5×100)/10

=50% Ans.

By Tricks:

=((15-10)/10)×100

=(5/10)×100

=50% Ans.

(Q2)The selling price of 12 objects is equal to the cost price of 9 objects. Find the Loss or Profit percentage?

Solution: By Concept And Shortcut Tricks

S.P of 12 objects =C.P of 9 objects

S.P/C.P=9/12

Note:C.P>S.P so loss

Loss=C.P-S.P

=12-9=3

%Loss=(3×100)/12

=25% Ans.

By Tricks:

=((12-9)/12)×100

=(3/12)×100

=25% Ans.

(Q3)By selling 33m cloth a shopkeeper earns profit equivalent to selling price of 11m. Find his profit percent.

Solution: By Concept And Shortcut Tricks

Profit     =    S.P-C.P

S.P(11m)=s.p(33m)-C.P(33m)

⇒C.P(33m)=s.p(33m)-S.P(11m)

⇒C.P(33m)=s.p(22m)

⇒C.P/S.P=22m/33m

Profit=33m-22m=11m

% Profit=(11×100)/22

=50% Ans

By Tricks:

% Profit=(Profit/S.P-Profit)×100

=(11/33-11)×100

=(11/22)×100

=50% Ans.

(Q4)By selling 33m, a shopkeeper incurs loss equivalent to the cost price of 11m. Find his loss percent.

Solution: By Concept And Shortcut Tricks

Loss     = C.P-S.P

C.P(11m)=C.P(33m)-S.P(33m)

⇒S.P(33m)=C.P(33m)-C.P(11m)

⇒S.P(33m)=C.P(22m)

⇒S.P/C.P=22m/33m

Loss=(33-22)m=11m

%Loss=(11×100)/33

=33 ¹/3% Ans.

The common and frequently asked a question of Profit and Loss Problem with solution and Shortcut Tricks for Bank PO, SSC CGL, SSC CHSL, Railway Exam Type-3.2

If two article are sold at equal prices, the First one is sold at X% profit whereas the second one at a loss of X% then the seller always incurs a loss in the entire Transition.

It is equal to ( X)²/100

(Q1) A dealer sold two T.V sets for Rs 7400 each on one he gained 10% and on the other, he lost 10%. The dealer’s loss or gain in the transaction is.

Solution:

% Loss=( X)²/100

=( 10)²/100

=100/100=1% Ans.

(Q2) a dealer sold two T.V sets for Rs 990 each on one he gained 10% and on the other he lost 10%. Then

(1)Find total C.P of T.V.

(2)His loss on the whole in Rs is.

Solution:

% Loss=( X)²/100

=( 10)²/100

=100/100=1%

x×(100-1)/100=1980

x×99/100=1980

∴ x=(1980×100)/99

=Rs 2000 Ans.

(2)Loss=C.P-S.P

=Rs(2000-1980)

=Rs 20 Ans.

The common and frequently asked a question of Profit and Loss Problem with solution and Shortcut Tricks for Bank PO, SSC CGL, SSC CHSL, Railway Exam Type-3.3

Trick: If two article is bought at equal prices, the First one is sold at X% profit whereas the second one at a loss of X% then the seller always incurs a no loss and no profit in the entire Transition.

(Q1)A dealer bought two T.V sets for Rs 7400 each. The First one is sold at 10% profit whereas the second one at a loss of 10%. Then find out loss or gain in the transaction.

Solution: No Loss And No Profit.

Note:[We know that If two article is bought at an equal price, the First one is sold at X% profit whereas the second one at a loss of X% then the seller always incurs a no loss and no profit in the entire Transition.]

The common and frequently asked a question of Profit and Loss Problem with solution and Shortcut Tricks for Bank PO, SSC CGL, SSC CHSL, Railway Exam Type-3.4

Trick: If two article are sold at equal prices, the First one is sold at P% profit whereas the second one at a loss of L% then the

Note: If +ve then profit or if -ve then loss

(Q1) A man sold two watches for Rs 525 each. On one he gains 10% and on the other he loses 15%. His profit or loss in the whole transaction is.

Solution:

=((100×-5)-2×10×15)/110+85

=(-500-300)/195

=-4.10% ie Loss=4.10% Ans.

By Concept:

100x×(100+10)/100=525

⇒ (100x×110)/100=525

∴ x=525/110

C.P of 1st watch=100x

=100×525/110

=Rs 477.27

Now,

100x×(100-15)/100=525

⇒ (100x×85)/100=525

∴ x=525/85=6.17

C.P of 2nd watch=100x

=100×6.17

=Rs 617

Now,

Total C.P=Rs(477.27+617)

=Rs 1094.27=1094

Total S.P=525+525=Rs 1050

Loss=C.P-S.P

=Rs(1094-1050)

=Rs 48

% Loss=(48×100)1094

=4.10% Ans.

TRICK-1

If two article are sold at equal prices, the first one is sold at P1% Profit and whereas the second one at a Profit of P2% And the sum of the cost price of two article is x then

For Example:

(Q1) A trader bought two watches for Rs 2300. He sold one at a profit of 10% and the other at a profit of 20%. If the selling price of each watch is the same, then their cost price are respectively.

Solution: By Trick:

C.P of watch at a profit of 10%=((100+20)/(200+10+20))×2300

=(120/230)×2300

=Rs 1200 Ans.

C.P of watch at a profit of 20%=Rs(2300-1200)

=Rs 1100 Ans.

BY CONCEPT:

Let C.P of 1st watch=x

C.P of 2nd watch=2300-x

S.P of 1st watch:

x×(100+10)/100=S.P1

11x/10=S.P1

S.P of 2nd watch:

(2300-x)×(100+20)/100=S.P2

(2300-x)×120/100=S.P2

(2300-x)×12/10=S.P2

Now According to question: selling price of each watch is the same

SO,

S.P1=S.P2

11x/10=(2300-x)×12/10

11x=2300×12-12x

23x=2300×12

x=Rs 1200 Ans

C.P of 2nd watch=2300-x

=Rs(2300-1200)

=Rs 1100 Ans.

TRICK-2

If two article are sold at equal prices, the first one is sold at a Loss of L1% and whereas the second one at a Loss of L2%, And the sum of the cost price of two article is x then

For Example:

(Q1) A Man purchases two T.V sets for Rs 21,500. He sold one at a Loss of 12% and the other at a loss of  16%. If the selling price of each T.V is the same, then their cost price are respectively.

Solution: By Trick:

C.P of T.V at a Loss of 12%=((100-16)/(200-12-16))×21500

=(84/172)×21500

=Rs 10500 Ans.

C.P of T.V at a Loss of 16%=Rs(21500-10500)

=Rs 11000 Ans.

BY CONCEPT:

Let C.P of 1st T.V =x

C.P of 2nd T.V=21500-x

S.P of 1st T.V:

x×(100-12)/100=S.P1

88x/100=S.P1

S.P of 2nd T.V:

(21500-x)×(100-16)/100=S.P2

(21500-x)×84/100=S.P2

(21500-x)×84/100=S.P2

Now According to question: selling price of each T.V is the same

SO,

S.P1=S.P2

88x/100=(21500-x)×84/100

88x=21500×84-84x

172x=21500×84

x=Rs 1o500 Ans

C.P of 2nd T.V=21500-x

=Rs(21500-10500)

=Rs 11000 Ans.

TRICK-3

If two article are sold at equal prices, the first one is sold at a Profit of P% and whereas the second one at a Loss of L%, And the sum of the cost price of two article is x then

For Example:

(Q1) A Man purchases two horses for Rs 16,800. He sold one at a Profit of 25% and the other at a loss of  15%. If the selling price of each horse is the same, then their cost price are respectively.

Solution: By Trick:

C.P of horse at a profit of 25%=((100-15)/(200+25-15))×16800

=(85/210)×16800

=(85/210)×16800

=Rs 6800 Ans.

C.P of horse at a Loss of 15%=Rs(16800-6800)

= Rs 10000 Ans.

BY CONCEPT:

Let C.P of 1st horse=x

C.P of 2nd horse=16800-x

S.P of 1st horse:

x×(100+25)/100=S.P1

125x/100=S.P1

S.P of 2nd horse:

(16800-x)×(100-15)/100=S.P2

(16800-x)×85/100=S.P2

(16800-x)×85/100=S.P2

Now According to question: selling price of each horse is the same

SO,

S.P1=S.P2

125x/100=(16800-x)×85/100

125x=16800×85-85x

210x=16800×85

x=Rs 6800 Ans

C.P of 2nd horse=16800-x

=Rs(16800-6800)

=Rs 10000 Ans.

TRICK-4

If two article are sold at equal prices, the first one is sold at a Loss of L% and whereas the second one at a Profit of P%, And the sum of the cost price of two article is x then

For Example:

(Q1) A trader bought two horses for Rs 19,500. He sold one at a Loss of 20% and the other at a Profit of  15%. If the selling price of each horse is the same, then their cost price are respectively.

Solution: By Trick:

C.P of horse at a Loss of 20%=((100+15)/(200-20+15))×19500

=(115/195)×19500

=Rs 11500 Ans.

C.P of horse at a Profit of 15%=Rs(19500-11500)

=Rs 8000 Ans.

The common and frequently asked problem of Profit and Loss question with solution and Shortcut Tricks for Bank PO, SSC CGL, SSC CHSL, Railway Exam Type-4.1.

(Q1)The total cost price of two watches is Rs 840. One is sold at a profit of 16% and the other at a loss of 12%. There is no loss or gain in the whole transition. Then their cost price is respectively.

Solution: BY CONCEPT

Let C.P of 1st watch=x

C.P of 2nd watch=840-x

S.P of 1st watch:

x×(100+16)/100=S.P1

116x/100=S.P1

S.P of 2nd watch:

(840-x)×(100-12)/100=S.P2

(840-x)×88/100=S.P2

Now According to question: There is no profit or loss

SO,

S.P1+S.P2=840

116x/100+(840-x)×88/100=840

116x+840×88-88x=840×100

28x=84000-73920

28x=10080

∴ x=360 Ans.

C.P of 2nd watch=840-x

=Rs(840-360)

=Rs 480 Ans.

2nd Method:

C.P of 1st watch at a profit 0f 16%=(3/7)×840

=Rs 360 Ans.

C.P of 2nd watch at a Loss of 12%=(4/7)×840

=Rs 480 Ans

OR

C.P of 2nd watch at a Loss of 12%=Rs(840-360)

=Rs 480 Ans.

(Q2) A man bought a horse and carriage for Rs 40000. He sold the horse at a gain of 10% and the carriage at a loss of 5%. He gained 1% on his whole transaction. The cost price of the horse was.

Solution: BY CONCEPT

Let C.P of horse=x

C.P of a carriage=40000-x

S.P of a horse:

x×(100+10)/100=S.P

11x/10=S.P

S.P of a carriage:

(40000-x)×95/100=S.P

Now According to question: he gained 1% on his whole transaction

So,

S.P of a horse+S.P of a carriage=40000×101/100

11x/10+(40000-x)×95/100=40000×101/100

11ox+40000×95-95x=40000×101

15x=40000×101-40000×95

=40000(101-95)

15x=240000

∴ x=Rs 16000 Ans

2nd Method:

C.P of a horse at a profit 0f 10%=(2/5)×40000

=Rs 16000 Ans.

(Q1)By selling a chair for Rs 350 instead of Rs 400, loss percent increases by 5%. The cost price of the chair is.

Solution: By Trick:

C.P=((400-350)/5)×100

=(50/5)×100

=Rs 1000 Ans.

BY CONCEPT:

S.P1=Rs 350

S.P2=Rs 400

Let C.P=x

Loss1(L1)=x-350

Loss2(L2)=x-400

%L1=((x-350)×100)/x

Note: % loss=(loss×100)/C.P

%L2=((x-400)×100)/x

Now,

L%=L1% – L2%

5=((x-350)×100)/x – ((x-400)×100)/x

5=100/x(x-350 – x-400)

5=(100/x)×50

5x=100×50

x= Rs 1000 Ans.

(Q2)5% more is gained by selling a cow for Rs 1010 than by selling it for Rs 1000. Find the cost price of the cow.

Solution: By Trick:

C.P=((1010-1000)/5)×100

=(10/5)×100

=Rs 200 Ans.

BY CONCEPT:

S.P1=Rs 1010

S.P2=Rs 1000

Let C.P=x

Profit1(P1)=1010-x

Profit2(P2)=1000-x

%P1=((1010-x)×100)/x

Note: % Profit=(Profit×100)/C.P

%P2=((1000-x)×100)/x

Now,

P%=P1% – P2%

5=((1010-x)×100)/x – ((1000-x)×100)/x

5=100/x(1010-x – 1000-x)

5=(100/x)×10

5x=100×10

x= Rs 200 Ans.

(Q3) A loss of 19% gets converted into a profit of 17% when the selling price is increased by Rs 162. The cost price of the article is.

Solution: By Concept:

[Concept: according to question 19% के loss पर जो selling price है  उसमे Rs 162 increased करने पर loss convert हो गया 17% profit में so 81x+162=117x]

Now,

81x+162=117x

36x=162

∴ x=4.5

C.P=100x=100×4.5

=Rs 450 Ans.

BY TRICK:

C.P=(162×100/(17+19)

=(162×100)/36

=Rs 450 Ans.

The common and frequently asked a question of Profit and Loss problem with solution and Shortcut Tricks for Bank PO, SSC CGL, SSC CHSL, Railway Exam Type-5.1

TRICK:

The profit obtained by selling an article for Rs X is same as the loss incurred if it is sold for Rs Y then

For Example:

(Q1)The profit obtained by selling an article for Rs 340 is same as the loss incurred if it is sold for Rs 248. Find the C.P.

Solution: By Concept

Let C.P=x

Profit=340-x    [Profit=S.P-C.P]

Loss=x-248      [Loss=C.P-S.P]

Now, According to Question

340-x=x-248

-2x=-588

∴ x= Rs 294

So, C.P=Rs 294 Ans

By Trick:

C.P=(340+248)/2

=Rs 294 Ans

(Q2)The profit obtained by selling a table for Rs 625 is same as the loss incurred if it is sold for Rs 545. The price at which it is to be sold to realize a profit of Rs 65 on the cost price is.

Solution: By Concept

Let C.P=x

Profit=625-x    [Profit=S.P-C.P]

Loss=x-545      [Loss=C.P-S.P]

Now, According to Question

625-x=x-545

2x=1170

x= Rs 585

∴ Required S.P=Rs(585+65)

= Rs 650 Ans

By Trick:

C.P=(625+545)/2

=Rs 585 Ans

∴ Required S.P=Rs(585+65)

= Rs 650 Ans

(Q3)The profit earned by selling an article for Rs 832 is equal to the loss incurred when the same article is sold for Rs 448. What should be the sale price for making 50% profit?

Solution: By Concept

Let C.P=x

Profit=832-x    [Profit=S.P-C.P]

Loss=x-448      [Loss=C.P-S.P]

Now, According to Question

832-x=x-448

2x=1280

x= Rs 640

Now,

640×(100+50)/100=S.P

640×(150/100)=S.P

S.P=Rs 960 Ans.

By Trick:

C.P=(832+448)/2

=Rs 640

Now,

640×(100+50)/100=S.P

640×(150/100)=S.P

S.P=Rs 960 Ans.

The common and frequently asked a question of Profit and Loss problem with solution and Shortcut Tricks for Bank PO, SSC CGL, SSC CHSL, Railway Exam Type-5.2.

TRICK:

The profit earned by selling an article for Rs X is n times the loss incurred when the same article is sold for Rs Y. Then

For Example:

(Q1)The profit earned by selling an article for Rs 900 is double the loss incurred when the same article is sold for Rs 450. At what price should the article be sold to make 25% profit?

Solution: By Concept

Let C.P=x

Profit=900-x    [Profit=S.P-C.P]

Loss=x-450      [Loss=C.P-S.P]

Now, According to Question

900-x=2(x-450)

3x=1800

x= Rs 600

Now,

600×(100+25)/100=S.P

600×(125/100)=S.P

S.P=Rs 750 Ans.

By Trick:

C.P=(900+2×450)/(2+1)

=1800/3

=Rs 600

Now,

600×(100+25)/100=S.P

600×(125/100)=S.P

S.P=Rs 750 Ans.

(Q2)Profit earned by selling an article for Rs 1060 is 20% more than the loss incurred by selling the article for Rs 950. At what price should the article be sold to earn 20% profit?

Solution: By Concept

Let C.P=x

Profit=1060-x    [Profit=S.P-C.P]

Loss=x-950      [Loss=C.P-S.P]

Now, According to Question

1060-x=(120/100)×(x-950)

5(1060-x)=6(x-950)

5300-5x=6x-5700

-11x=-11000

x= Rs 1000

Now,

1000×(100+20)/100=S.P

S.P=Rs 1200 Ans.

BY TRICK:

C.P=(1060+(120/100)×950)/(120/100)+1

=(1060+95×12)/(22/10)

=Rs 1000

Now,

1000×(100+20)/100=S.P

S.P=Rs 1200 Ans.

(Q3)A shopkeeper sells an article for Rs 140 and earns double the profit that he would have earned had he sold it for Rs 120. What is its cost price?

Solution: By trick

C.P=(2×120-140)/2-1

=Rs 100 Ans.

By Concept:

Let C.P=x

Profit=140-x    [Profit=S.P-C.P]

Profit=120-x

Now, According to Question

140-x=2×(120-x)

x=Rs 100 Ans.

(Q1)Successive discounts of 30% and 20% are equivalent to a single discount of.

Solution:

=44% Ans.

(Q2)A single discount equivalent to the successive discounts of 10%, 20%, and 25% is.

Solution:

=28%

Now,

=46% Ans.

(Q3)The difference between successive discounts of 40% followed by 30% and 45% followed by 20% on the marked price of an article is Rs 12. The marked price of the article is.

Solution:

=58%

=56%

Let M.P=100x

Difference=58%-56%=2%

According to the question

100x×(2/100)=12

so, x=6

∴ M.P=100x=100×6

=Rs 600 Ans.

(Q4)The marked price of an article is Rs 500. It is sold at successive discounts of 20% and 10%. The selling price of the article is.

Solution:

=28%

Now,

500×(100-28)/100=S.P

⇒  500×(72/100)=S.P

∴  S.P=Rs 360 Ans.

(Q5)A trader allows two successive discounts of 30% and 15% on selling an article. If he gets Rs 476 for that article, Find its marked price.

Solution:

=40.5%

Now,

100x×(100-40.5)/100=476

⇒ 100x×(59.5/100)=476

x=476/59.5

∴ M.P=100x=100×(476/59.5)

=Rs 800 Ans.

(Q6)The difference between a discount of 30% on Rs 2000 and two successive discounts of 25% and 5% on the same amount is.

Solution:

Discount=2000×(30/100)

= Rs 600

=28.75%

Now,

Discount=2000×(28.75/100)

= Rs 575

∴ Difference=Rs(600-575)

= Rs 25 Ans.

(Q7)A dealer buys a car listed at 200000 at successive discounts of 10% and 5%. If he sells the car for 179550, then his profit percent is.

Solution:

=14.5%

[Concept: suppose dealer company से कार खरीदता है| company car का marked price 2००००० रखता है or company car पर two successive discounts दे रहा है 10% and 5%. इससे company का selling price निकाल लेंगे, company का जो selling price होगा वह dealer का cost price होगा|]

200000×(100-14.5)/100=S.P

200000×(85.5/100)=S.P

S.P=Rs 171000

C.P of dealer=Rs 171000

S.P of dealer(Given)=Rs 179550

Gain=Rs(179550-171000)

=Rs 8550

% profit=(8550×100)/171000

=5% Ans

(Q8)A dealer buys a table listed at Rs 1500 and gets successive discounts of 20% and 10%. He spends Rs 20 on transportation and sells it at a profit of 20%. Find the selling price of the table.

Solution:

=28%

[Concept: suppose dealer company से Table खरीदता है| company Table का marked price 1500 रखता है or company Table पर two successive discounts दे रहा है 20% and 10%. इससे company का selling price निकाल लेंगे, company का जो selling price होगा वह dealer का cost price होगा|]

1500×(100-28)/100=S.P

1500×(72/100)=S.P

S.P=Rs 1080

C.P of dealer=Rs 1080

Total C.P of dealer=Rs(1080+20)

=Rs 1100

1100×(100+20)/100=S.P

1100×(120)/100=S.P

S.P=Rs 1320 Ans.

(Q9)The marked price of a watch is 1600. The shopkeeper gives successive discounts of 10% and x% to the customer. If the customer pays Rs 1224 for the watch. The value of x.

Solution:

1600×(100-10)/100=S.P

⇒1600×(90/100)=S.P

S.P=Rs 1440

Now,

(S.P after x% discount)

[Concept:C.P of customer=1224 i.e S.P of shopkeeper]

1440×(100-x)/100=1224

144000-1440x=122400

x=15% Ans.

(Q10)A bicycle marked at Rs 2000 is sold with two successive discount of 20% and 10%. An additional discount of 5% is offered for cash payment. The selling price of the bicycle at cash payment is.

Solution:

=28%

Now,

=31.6%

=2000×(100-31.6)/100=S.P

=2000×(68.4/100)=S.P

∴ S.P=Rs 1368 Ans.

(Q1)A person sells an article at a profit of 10%. If he had bought it at 10% less and sell it for Rs 3 more he would have gained 25%.(1) Find the cost price (2) Find S.P

Solution:

Now,

90x×(100+25)/100=110x+3

90x×(125)/100=110x+3

x=6/5

∴ C.P=100x=100×(6/5)

=Rs 120 Ans.

(2)S.P=110x=110×(6/5)

=Rs 132 Ans.

(Q2)A person sold a T.V at a gain of 15 %. Had he bought it for 25% less and sold it for Rs 600 less, he would have made a profit of 32%. The cost price of the T.V was.

Solution:

Now,

75x×(100+32)/100=115x-600

75x×(132/100)=115x-600

99x=115x-600

x=600/16

C.P=100x=100×(600/16)

=Rs 3750 Ans.

(Q3)A shopkeeper sold a watch at a loss of 20%. But if he could sell it at Rs 200 more, he could earn a profit of 5%. The cost price of the watch is.

Solution:

Now,

1oox×(100+5)/100=80x+200

105x=80x+200

x=8

∴ C.P of watch=100x=100×8

=Rs 800 Ans

The common and frequently asked of Profit and Loss question with solution and Shortcut Tricks for Bank PO, SSC CGL, SSC CHSL, Railway Exam Type-6.1.

If a trader professes(Profess-दावा करना) to sell his goods at cost price, but uses false weights, then

For Example:

(Q1) A shopkeeper sells goods at the cost price but uses 900gms instead of 1 Kg. Find his overall profit or loss percentage.

Solution: BY TRICK:

Error= True Weight – False Weight

= 1000g – 900g=100g

% Profit= (100g × 100)/900g

=11 ¹/9% Ans.

BY CONCEPT:

Let C.P of shopkeeper for 1g=Rs 1

C.P of shopkeeper for 1kg ie 1000g=Rs 1000

False C.P of shopkeeper for 900g=Rs 900

[Note: 1,2 and 3 are Steps]

% Profit=(100×100)/900

=11 ¹/9% Ans.

Concept:[दुकानदार ने आप से कहा कि जिस रेट पर मैंने समान(Goods) खरीदा है उसी रेट पर में आप को समान(Goods) दे रहा हूँ| Means दुकानदार ने यहाँ Rs 1000 में समान खरीदा है और आपको भी Rs 1000 में दे रहा है| लेकिन दुकानदार 1 kg समान(Goods) के बदले सिर्फ आपको 900g समान(Goods)दे रहा है| यहाँ समझने वाली बात यह है कि दुकानदार आपसे 1000g(1Kg) का पैसा लिया that means आप ने दुकानदार को Rs 1000 दिया (So S.P of Shopkeeper=Rs 1000) लेकिन उसने सिर्फ आपको 900g समान(Goods) दिया means दुकानदार ने आपको सिर्फ Rs 900 का समान दिया So, C.P of Shopkeeper=Rs 900. ∴ profit=(S.P-C.P)=Rs(1000-900)=Rs 100]

(Q2)A shopkeeper sells his goods at 20% profit and uses 900gms instead of 1 Kg. Find his overall profit or loss percentage.

Solution:

Let C.P of shopkeeper for 1g=Rs 1

C.P of shopkeeper for 1kg ie 1000g=Rs 1000

False C.P of shopkeeper for 900g=Rs 900

% Profit=(300×100)/900

=33 ¹/3% Ans.

Rough:

C.P=Rs 1000 and Profit=20%

S.P=1000×(100+20)/100

=Rs 1200

(Q3) A shopkeeper marks his good at 20% above the cost price and also allows a discount of 20% but uses 800gms instead of 1Kg. Find overall profit or loss percentage.

Solution:

Let C.P of shopkeeper for 1g=Rs 1

C.P of shopkeeper for 1kg ie 1000g=Rs 1000

False C.P of shopkeeper for 800g=Rs 800

% Profit=(160×100)/800

=20% Ans.

Rough:

1000×120/100

=Rs 1200

1200×80/100

=Rs 960

Note: Discount always marked price पर calculate किया जाता है or discount के बाद जो price आता है वह selling price होता है|

(Q4) A dealer sells goods at 6% loss but uses 14g instead of 16g. What is his percentage profit or loss?

Solution:

Let C.P of shopkeeper for 1g=Rs 1

C.P of shopkeeper for 16g=Rs 16

False C.P of shopkeeper for 14g=Rs 14

% Profit=(1.04×100)/14

=7.42% Ans.

(Q5) A shopkeeper cheats to the extent of 10% while buying as well as selling, by using false weights. His total gain is.

Solution:

=(12100-10000)/100

=21% Ans.

Other Methods:

=21% Ans.

(Q6)A shopkeeper cheats to the extent of 15% while buying and 20%while selling, by using false weights. His total gain is.

Solution:

=38% Ans.

Other methods:

=38% Ans.

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